Iron burns in oxygen to form iron oxide Fe3O4. Determine the mass of iron scale that is formed

Iron burns in oxygen to form iron oxide Fe3O4. Determine the mass of iron scale that is formed during the combustion of 47.6 g of iron. How many liters of oxygen is required for the iron to react completely.

Let’s implement the solution:
1. We write down the process according to the condition of the problem:
m = 47.6 g. X l. -? X g -?
3Fe + 2O2 = Fe3O4 – compounds, scale is released;
2. Calculation by formulas:
M (Fe) = 55.8 g / mol;
M (O2) = 32 g / mol;
M (Fe3O4) = 231.4 g / mol;
Y (Fe) = m / M = 47.6 / 55.8 = 0.85 mol. (substance in excess)
3. Proportions:
0.85 mol (Fe) – X mol (O2);
-3 mol – 2 mol from here, X mol (O2) = 0.85 * 2/3 = 0.57 mol, (the substance is in short supply).
Calculations are made for the substance in deficiency.
0.57 mol (O2) – X mol (Fe3O4);
-2 mol -1 mol from here, X mol (Fe3O4) = 0.57 * 1/2 = 0.28 mol.
4. Find the volume of O2, the mass of scale:
V (O2) = 0.57 * 22.4 = 12.77 L;
m (Fe3O4) = Y * M = 0.28 * 231.4 = 64.79 g.
Answer: for oxidation, oxygen with a volume of 12.77 liters is required, and iron oxide with a mass of 64.79 g is obtained.